Forbidden subgraphs implying the MIN-algorithm ...

Discrete Mathematics 256 (2002) 193 – 201

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Forbidden subgraphs implying the MIN-algorithm gives a maximum independent set  Jochen Haranta , Zden,ek Ryj0ac, ekb , Ingo Schiermeyerc;∗ a Department

of Mathematics, Technical University of Ilmenau, D-98684 Ilmenau, Germany of Mathematics, University of West Bohemia, 306 14 Pilsen, Czech Republic c Department of Mathematics, Freiberg University of Mining and Technology, D-09596 Freiberg, Germany

b Department

Received 1 July 1999; received in revised form 27 July 2000; accepted 21 December 2000

Abstract The well-known greedy algorithm MIN for 6nding a maximal independent set in a graph G is based on recursively removing the closed neighborhood of a vertex which has (in the currently existing graph) minimum degree. We give a forbidden induced subgraph condition under which algorithm MIN always results in 6nding a maximum independent set of G, and hence yields the exact value of the independence number of G in polynomial time. c 2002 Elsevier Science B.V. All rights reserved.


1. Introduction Throughout the paper, we consider only 6nite undirected graphs G = (V (G); E(G)) without loops and multiple edges. By NG (x) we denote the neighborhood of a vertex x ∈V (G), i.e., the set of all neighbors of x. We further denote by NG [x] = NG (x) ∪ {x} the closed neighborhood of x in G, by dG (x) = |NG (x)| the degree of x in G and by (G) = min{dG (x) | x ∈V (G)} the minimum degree of G. For a set M ⊂V (G), we

, No. 201=97=0407. Research supported by Grant GA CR Corresponding author. TU Bergakademic Freiberg, Inst. fur Theor. Mathemtik, Bernhard-von-Cotta-Str. 2, 09596 Freiberg, Germany. E-mail addresses: [email protected] (J. Harant), [email protected] (Z. Ryj0ac, ek), [email protected] (I. Schiermeyer). 

∗

c 2002 Elsevier Science B.V. All rights reserved. 0012-365X/02/$ - see front matter  PII: S 0 0 1 2 - 3 6 5 X ( 0 2 ) 0 0 5 7 1 - X

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denote by M G the induced subgraph of G on M and we set G − M = V (G)\M G . By (G) we denote the independence number of G, i.e., the size of a maximum (i.e. largest) independent set in G. If F1 ; : : : ; Fk are graphs, then we say that G is {F1 ; : : : ; Fk }-free if G does not contain a copy of any of the graphs F1 ; : : : ; Fk as an induced subgraph. For other terminology and notation not de6ned here, we refer to [1]. The well-known greedy algorithm MIN for 6nding a maximal independent set in a graph G [4] can be stated as follows: Algorithm MIN (Minimum degree). 1. H1 := G; i := 1; SMIN := ∅. 2. Choose a vertex vi ∈V (Hi ) such that dHi (vi ) = (Hi ) and set SMIN := SMIN ∪ {vi }; Hi+1 := Hi − NHi [vi ]. 3. If V (Hi+1 ) = ∅ then i := i + 1 and go to 2. 4. STOP. Obviously, the set SMIN , generated by Algorithm MIN, is a maximal (but not necessarily maximum) independent set in G, and hence (G)¿|SMIN |. Mahadev and Reed [3] considered the following (also greedy) algorithm for 6nding a maximal independent set in G, based on an ordering of the vertices of G according to their degrees in G. This algorithm can be equivalently formulated as follows. Algorithm VO (Vertex order). 1. Order the vertices of G into a sequence v1 ; : : : ; vn such that dG (vj )6dG (vk ) for any j; k, 16j¡k6n. 2. G1 := G; i := 1; SVO := ∅. 3. For i := 1 to n do: If NG (vi ) ∩ SVO = ∅, then SVO := SVO ∪ {vi }. 4. STOP. It is clear that the set SVO , generated by Algorithm VO, is a maximal independent set in G, and hence also (G)¿|SVO |. Note that both Algorithm MIN and Algorithm VO have polynomial time complexity whereas the determination of (G) is diLcult since the corresponding decision problem INDEPENDENT SET is a well-known NP-complete problem [2]. Denote by k MIN (G) and kVO (G) the smallest cardinality of an independent set of G that Algorithm MIN and Algorithm VO can create, respectively. Let F1 ; : : : ; F6 be the graphs in Fig. 1 and let FA = {F1 ; F2 ; F3 ; F4 ; F5 ; F6 }. The following theorem, which forms the essential part of the main result of [3], shows that in the class of FA -free graphs, Algorithm VO always yields a maximum independent set. Theorem A (Mahadev and Reed [3]). Let G be an FA -free graph. Then kVO (G) = (G):

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Fig. 1.

Fig. 2.

2. Main result Let F7 ; : : : ; F13 be the graphs shown in Fig. 2 and let F1 = {F1 ; F3 ; F5 ; F6 ; F7 ; F8 ; F9 ; F10 ; F11 ; F12 ; F13 }. Since F2 is an induced subgraph of F7 , and F4 is an induced subgraph of each of the graphs F8 ; : : : ; F13 , the class of FA -free graphs is a proper subclass of the class of F1 -free graphs. Thus, the following theorem, which is the main result of this paper, extends Theorem A in the sense that even for F1 -free graphs the independence number can be calculated in polynomial time. Theorem 1. Let G be an F1 -free graph of order n¿7. Then k MIN (G) = (G):

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Equivalently, Theorem 1 gives a collection of forbidden induced subgraphs which imply that Algorithm MIN always yields a maximum independent set. The proof of Theorem 1 is postponed to Section 3. As already noted, FA -free ⇒ F1 -free. However, the price for a more general result is paid here in larger number of forbidden subgraphs. The following corollary of Theorem 1 avoids this drawback and still extends Theorem A. Let F2 = {F1 ; F3 ; F4 ; F5 ; F6 ; F7 }. Note that, since F7 contains an induced F2 and each of the graphs F8 ; : : : ; F13 contains an induced F4 , we have FA -free ⇒ F2 -free ⇒ F1 -free. Corollary 2. Let G be an F2 -free graph of order n¿7. Then k MIN (G) = (G): The following statement shows that Corollary 2 (and hence also Theorem 1) is considerably stronger than Theorem A. More speci6cally, it says that under the assumptions of Corollary 2, the diOerence between the output of Algorithm MIN and that of Algorithm VO can be arbitrarily large. Theorem 3. For every integer k there is an F2 -free graph G such that k MIN (G) − kVO (G)¿k: Proof. Let G be the class of graphs de6ned recursively as follows: (i) F2 ∈G, (ii) for any G1 ; G2 ∈G, let also (G1 + G2 ) ∨ K1 ∈G and (G1 + G2 ) ∨ K2 ∈G. (Following [1], we denote by “+” the disjoint union and by “∨” the join of two graphs, respectively.) We show that every graph G ∈G is F2 -free. We 6rst have the following observation, the proof of which is obvious. Claim. Let F ∈F2 with |V (F)| = r. Then dF (x)6r − 2 for every x ∈V (F) and min{dF (x); dF (y)}6r − 3 for any pair of independent vertices x; y ∈V (F). Since F2 ∈F = 2 , the graph F2 is F2 -free. Suppose now that G1 ; G2 are F2 -free. If (G1 + G2 ) ∨ K1 or (G1 + G2 ) ∨ K2 contains an induced F ∈F2 , then, since F is connected, V (F) contains at least one vertex outside V (G1 ) ∪ V (G2 ), but then we have a contradiction with the claim. Hence, every graph in G is F2 -free.
If we now set G1
= F2 and Gi+1 = (Gi
+ Gi
) ∨ K1 for i¿1, then Gi
∈G for any i¿1
and it is apparent that k MIN (Gi ) = (Gi
) = 3 · 2i−1 , but kVO (Gi
) = 2i . Remark. By Theorem 1, in the class of F1 -free graphs, Algorithm MIN is always at least as good as Algorithm VO and by Theorem 3 the diOerence can be arbitrarily large. The following construction shows that without the assumption of F1 -freeness

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Algorithm VO can be better than Algorithm MIN (i.e., for all graphs, the two algorithms are incomparable). Let p¿3 be an arbitrary integer, let G 1  G 2  Kp , G 3  K1 and G 4  Kp be vertexdisjoint, let Gp
be the graph obtained by joining by an edge all pairs of vertices x; y for x ∈V (G i ), y ∈V (G i+1 ) (mod 4), and let Gp be the graph obtained by adding one new vertex to Gp
and joining it to all vertices of G 2 . Then clearly k MIN (Gp ) = 3, while kVO (Gp ) = p + 1. Since Algorithm MIN is (clearly) polynomial, we further have the following consequence of Theorem 1. Corollary 4. In the class of F1 -free graphs, the independence number can be computed in polynomial time. Note that it is obvious that F1 -free graphs are recognizable in polynomial time.

3. Proof of Theorem 1 We basically follow the general idea of the proof of Theorem A in [3], by replacing Algorithm VO with Algorithm MIN and the set FA by the set F1 . For the sake of clarity, whenever we list vertices of some induced subgraph F, we always order the vertices of the list such that their degrees (in F) form a nonincreasing sequence (with the exception of F1  P7 , where the ordering follows the path). Let G be a (without loss of generality) connected graph satisfying the assumptions of Theorem 1 and suppose that Algorithm MIN creates a maximal independent set S in G such that |S| = m¡(G), i.e., such that S is not maximum. Let the notation of vi ; Hi be chosen in accordance with the description of Algorithm MIN in Section 1, i.e., such that S = {v1 ; : : : ; vm }, H1 = G, dHi (vi ) = (Hi ) and Hi+1 = Hi − N [vi ], and set Sj = S ∩ V (Hj ) = {vj ; : : : ; vm }; j = 1; : : : ; m. Choose a maximum independent set T = {t1 ; : : : ; t } in G such that |S ∩ T | is maximum, and set Tj = T ∩ V (Hj ); j = 1; : : : ; m. Since both S and T are independent, S ∪ T G is bipartite with all its isolated vertices in S ∩ T . Let R be a component of S ∪ T G with |R ∩ S|¡|R ∩ T | (such an R always exists since |S|¡|T |) and set k = min{i ∈ {1; : : : ; m} | vi ∈R ∩ S} (with a slight abuse of notation, we will use R for both the component and its vertex set). We have the following observations. Claim 1. Sj is a dominating set in Hj , j = 1; : : : ; m. Proof. If x ∈V (Hj )\Sj , then NG (x) ∩ {v1 ; : : : ; vj−1 } = ∅, since otherwise x ∈V = (Hj ) by the de6nition of Hj . Since S is a dominating set in G, necessarily NG (x) ∩ Sj = ∅, implying NHj (x) ∩ Sj = ∅. Claim 2. dHj (x)¿dHj (vj ) for every x ∈V (Hj ) and for every j = 1; : : : ; m.

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Proof. Follows immediately from the de6nition of Algorithm MIN. Claim 3. R ⊂ V (Hk ). Proof. Obviously, R ∩ S ⊂ V (Hk ). If y ∈(R ∩ T )\V (Hk ), then y ∈NHj (vj ) for some j¡k and hence vj ∈R ∩ S, contradicting the choice of k. Hence also R ∩ T ⊂ V (Hk ). The following simple observation will be often used implicitly throughout the proof. Claim 4. If F is a subgraph of Hj for some j ∈ {1; : : : ; m}, then F is induced in Hj if and only if F is induced in G. In the sequel, we will use the following notation: |R ∩ S| = p, |R ∩ T | = q, R ∩ S = {vi1 ; : : : ; vip }, R ∩ T = {t1 ; : : : ; tq }, and we suppose the notation of the vertices in R ∩ S is chosen such that i1 = k and ij1 ¡ij2 for j1 ¡j2 . Case 1: R contains a cycle. If R contains an induced cycle of length ‘¿8, then R contains also an induced F1 , a contradiction. Suppose that R contains an induced cycle C of length 6, and let C = s1 t1 s2 t2 s3 t3 s1 , where si ∈R ∩ S, ti ∈R ∩ T , i = 1; 2; 3. Since |R ∩ S|¡|R ∩ T |, there is a t4 ∈R ∩ T , adjacent to (say) s1 . If s2 t4 ∈E(G), then (since C is induced and T is independent), {s2 ; t4 ; s1 ; t1 ; t2 }G  F3 , a contradiction. Hence s2 t4 ∈E(G), = and similarly s3 t4 ∈E(G). = But then {s1 ; t1 ; s2 ; t2 ; s3 ; t3 ; t4 }G  F7 , a contradiction. Hence every induced cycle in R has length exactly 4. Since R is bipartite and F3 -free, it follows easily (by induction, starting with a C4 ) that R is a complete bipartite graph with 26|R ∩ S|¡|R ∩ T |. Consider the vertex vk ∈R ∩ S. We have dR (vk )¿dR (y) for every y ∈R ∩ T (since |R ∩ S|¡|R ∩ T |), but, on the other hand, by the choice of vk and by Claim 2, dHk (vk )6 dHk (y) for every y ∈R ∩ T . It follows that there are vertices z ∈V (Hk )\R and y ∈R ∩ T such that zy ∈E(G), but zvk ∈E(G). = Claim 5. Let z ∈V (Hk )\R be such that zvk ∈E(G) = and NR∩T (z) = ∅. Then NR∩S (z) = ∅ and NR∩T (z) = {t1 ; : : : ; tq }. Proof. Let (without loss of generality) zt1 ∈ E(G). Suppose 6rst that NR∩S (z) = ∅. Then NR (z) = R ∩ T , since otherwise R ∪ {z}Hk contains an induced F3 . Since S is dominating, zs ∈E(G) for some s ∈S\R. Then NR∩S (s) = ∅ (since S is independent) and NR∩T (s) = ∅ (otherwise s ∈R), implying NR (s) = ∅ and {z; t1 ; vk ; t2 ; s}G  F3 . Hence NR∩S (z) = ∅. Let (without loss of generality) zvi2 ∈E(G). Recall that i1 = k; i.e., vi1 = vk . If zta ; ztb ∈= E(G) for some a; b ∈ {2; : : : ; q}, then {vi2 ; ta ; vi1 ; tb ; z}G  F3 , and if zta ∈E(G) = and ztb ∈E(G) for some a; b ∈ {2; : : : ; q}, then {vi1 ; t1 ; z; tb ; ta }G  F3 . Hence zti ∈E(G) for every i = 1; : : : ; q. Now, by Claim 5, q¿4 implies {z; vi2 ; vi1 ; t1 ; t2 ; t3 ; t4 }G  F8 . Hence q = 3 and, consequently, p = 2.

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Denote H = {z; vi2 ; vi1 ; t1 ; t2 ; t3 }G (note that H  F4 ). Since |V (G)|¿7, there is a vertex y ∈V (G)\V (H ) with NH (y) = ∅. Suppose 6rst that yvi1 ∈E(G). If y ∈V (Hk ) and yti ∈E(G) for i = 1; 2; 3, then {y; z; vi1 ; vi2 ; t1 ; t2 ; t3 }G is isomorphic to one of the graphs F11 , F12 or F13 , depending on the = then, by the choice of existence of the edges yvi2 ; yz. If y ∈V (Hk ) and (say) yt1 ∈E(G), vk (as a vertex of minimum degree in Hk ) and by Claim 2, there is a z
∈V (Hk )\V (H ) such that z
vi1 ∈E(G), = but z
t1 ∈E(G). By Claim 5, {vi2 ; t1 ; t2 ; t3 } ⊂ NH (z
), i.e., V (H )\{z} ∪ {z
}G  F4 . Then V (H ) ∪ {z
}G induces F10 or F9 , depending on whether zz
∈E(G) or not. If y ∈V = (Hk ), then yvi 0 ∈E(G) for some i0 , 16i0 ¡k. Note that NH (vi 0 ) = ∅ (since i0 ¡k). Then either NH (y) = V (H ), implying V (H ) ∪ {y}G  F11 , or y is nonadjacent to some vertex of H , and then it is easy to see that V (H ) ∪ {y; vi 0 }G contains an induced F3 for any possible structure of NH (y). This contradiction proves that = yvi1 ∈E(G). If NR∩T (y) = ∅, then yz ∈E(G) or yvi2 ∈E(G), but in both cases we have an induced F3 . Hence, NR∩T (y) = ∅. By Claim 5, yvi2 ∈E(G) and yti ∈E(G) for i = 1; 2; 3. Then again V (H ) ∪ {y} induces an F10 or F9 , depending on whether yz ∈E(G) or not. This contradiction completes the proof in Case 1. Case 2: R is a tree. Claim 6. All leaves of R are in T . Proof. If s ∈S is a leaf of R and t ∈T is the (only) neighbor of s in R, then T \{t} ∪ {s} is also a maximum independent set, contradicting the maximality of |S ∩ T |. Claim 6 immediately implies that every longest path in R has an odd number of vertices. Since G is F1 -free, a longest path in R can be only a P3 or a P5 . Subcase 2.1: R contains a P3 , both endvertices of which are leaves of R. By Claim 6, let ta vi‘ tb (where 16‘6p and 16a; b6q) be the vertices of the P3 . = (Hi‘ ) would imply ta vc ∈ First observe that ta ; tb ∈V (Hi‘ ) (since otherwise e.g. ta ∈V E(G) for some c, 16c¡i‘ , but then for 16c¡k the vertex ta would not be in Hk , and for k6c¡i‘ the vertex ta would not be a leaf of R). By Claim 2, there are vertices xa ; xb ∈V (Hi‘ )\R such that xa ta ∈E(G) and xb tb ∈E(G), but xa vi‘ ; xb vi‘ ∈E(G). = By Claim 1, each of xa , xb has a neighbor (say, va
and vb
) in Si‘ . Note that va
; vb
are nonadjacent to ta and tb (otherwise ta , tb are not leaves). Now we have xa = xb (otherwise {xa ; ta ; vi‘ ; tb ; va
}G  F3 ), xa tb ∈E(G) = and xb ta ∈E(G) = (otherwise = and va
{xa ; tb ; vi‘ ; ta ; va
}G  F3 or {xb ; ta ; vi‘ ; tb ; vb
}G  F3 ) and, 6nally, xa xb ∈E(G) = vb
(otherwise {xa ; xb ; ta ; tb ; vi‘ ; va
; vb
}G induces F1 , F6 or F5 ). Since the vertex va
( = vb
) is in Si‘ (but not necessarily in R), we have va
= v‘
for some ‘
, i‘ ¡‘
6m. Suppose that, among all common neighbors of xa , xb in Si‘ , va
is chosen such that ‘
is minimum. Then xa ; xb ∈H‘
, but ta ; tb ∈H = ‘
. By Claim 2 (for j = ‘
), there are za ; zb ∈V (H‘
) such that za xa ∈E(G) and zb xb ∈E(G), but za ; zb ∈N = G (v‘
) (and also za ; zb ∈N = G (vi‘ )). Suppose 6rst that za = zb . Since {xa ; za ; xb ; v‘
; ta }G  F3 , we have ta za ∈E(G). Symmetrically, {xb ; za ; xa ; v‘
; tb }G  F3 implies tb za ∈E(G). Then za ∈S = ‘
(otherwise ta ; tb

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are not leaves). By Claim 1, za has a neighbor s in S, but then {za ; ta ; vi‘ ; tb ; s}G  F3 . Hence, za = zb (implying za xb ∈E(G) = and zb xa ∈E(G)). = We show that za ta ∈E(G). = Let za ta ∈E(G). If za tb ∈E(G), then {tb ; vi‘ ; ta ; za ; xb }G  = Clearly za ∈S = (otherwise ta is not a leaf) and hence, by Claim 1, F3 ; hence za tb ∈E(G). za has a neighbor sa in S. Obviously, sa is not adjacent to any of vi‘ ; v‘
; ta ; tb . If sa xb ∈E(G), = then {sa ; za ; ta ; vi‘ ; tb ; xb ; v‘
}G  F1 ; hence sa xb ∈E(G), but then for sa xa ∈ E(G) we have {xb ; v‘
; xa ; sa ; tb }G  F3 , and for sa xa ∈E(G) = we have {xa ; za ; sa ; xb ; = v‘
; ta }G  F5 . Hence za ta ∈E(G). Since {xa ; ta ; vi‘ ; tb ; xb ; v‘
; za }G  F7 , we obtain za tb ∈E(G). Symmetrically, zb tb ∈= E(G) and zb ta ∈E(G). This also implies that za ; zb ∈S = (otherwise ta or tb is not a leaf). By Claim 1, there are vertices sa ; sb ∈S‘
such that za sa ∈E(G) and zb sb ∈E(G) (possibly sa = sb ). Obviously, sa and sb are not adjacent to any of ta ; tb ; vi‘ ; v‘
. If sa xa ∈E(G), then for sa = sb and xb sb ∈E(G) we have {xa ; v‘
; xb ; sa ; ta }G  F3 , other= and, similarly, sa xb ∈E(G). = But wise {xa ; ta ; vi‘ ; tb ; xb ; v‘
; sa }G  F7 . Hence sa xa ∈E(G) then {xa ; za ; tb ; xb ; v‘
; ta ; sa }G  F6 . This contradiction completes the proof in Subcase 2.1. Subcase 2.2: R contains no P3 both endvertices of which are leaves of R. In this subcase, R contains a P5 (but no P7 ). Using Claim 6, it is easy to show (by induction, starting with a P5 ) that R is isomorphic to the subdivision of a star with center and leaves in R ∩ T and with vertices of degree 2 in R ∩ S. Choose the notation such that vij is adjacent to the center t0 and to the leaf tj , j = 1; : : : ; p. By = Clearly, Claim 2, there is a vertex z1 ∈V (Hi1 ) such that z1 t1 ∈E(G), but z1 vi1 ∈E(G). z1 ∈S; = thus, by Claim 1, z1 v‘ ∈E(G) for some v‘ ∈S with i1 ¡‘6m. Note that v‘ is not adjacent to any of vi1 , t1 , but possibly t0 v‘ ∈E(G). Suppose 6rst that z1 tj ∈E(G) for some j, 26j6p. Then clearly also v‘ vij ; v‘ tj ∈= E(G). We further have t0 z1 ∈E(G) = (since otherwise {z1 ; t1 ; vi1 ; t0 ; tj }G  F3 ) and vij z1 ∈= E(G) (otherwise {vij ; z1 ; t1 ; vi1 ; t0 ; tj }G  F5 ). Now we have t0 v‘ ∈E(G), since otherwise {z1 ; t1 ; vi1 ; t0 ; vij ; tj ; v‘ }G  F7 . This implies v‘ ∈R ∩ S and, by the structure of R, v‘ has a (unique) neighbor t‘ in R ∩ T . But now {t‘ ; v‘ ; z1 ; t1 ; vi1 ; t0 ; vij }G  F6 if t‘ z1 ∈E(G), = or {v‘ ; z1 ; t1 ; vi1 ; t0 ; t‘ }G  F5 , if t‘ z1 ∈E(G), respectively. This contradiction proves that z1 is not adjacent to any of t2 ; : : : ; tp . Now suppose that z1 via ∈E(G) = for some a, 26a6p. Then t0 z1 ∈E(G) = (otherwise {t0 ; vi1 ; t1 ; z1 ; via }G  F3 ) and t0 v‘ ∈E(G) (otherwise {v‘ ; z1 ; t1 ; vi1 ; t0 ; via ; ta }G  F1 ). This implies, as before, that v‘ is in R ∩ S and has a (unique) neighbor t‘ in R ∩ T , but then {v‘ ; t0 ; vi1 ; t1 ; z1 ; via ; t‘ }G  F6 . Hence, z1 is adjacent to all vertices in (R ∩ S)\ {vi1 }. This immediately implies p = |R ∩ S| = 2, for otherwise {t0 ; vi2 ; z1 ; vi3 ; vi1 }G  F3 . Summarizing, it remains to consider the case when R ∩ S = {vi1 ; vi2 }, R ∩ T = {t0 ; t1 ; t2 } and NR (z1 ) = {t1 ; vi2 }. We consider the graph Hi2 . Since i1 ¡i2 , {vi1 ; t0 ; t1 } ∩ = we have z1 ; t2 ∈V (Hi2 ). By Claim 2 (for j = i2 ), V (Hi2 ) = ∅, and since z1 vi1 ; t2 vi1 ∈E(G), there is a vertex z1
∈V (Hi2 ) such that z1
z1 ∈E(G) but z1
vi2 ∈E(G) = (and, of course, z1
vi1 ∈E(G)). = If t0 z1
∈E(G), = then for t1 z1
∈E(G) we have {t1 ; z1 ; vi1 ; t0 ; vi2 ; z1
}G  F5 , and for t1 z1
∈E(G) = we have {z1 ; t1 ; vi1 ; t0 ; vi2 ; z1
; t2 }G  F6 if z1
t2 ∈E(G) = and
{vi2 ; t2 ; z1 ; z1 ; t0 }G  F3 if z1
t2 ∈E(G). Hence t0 z1
∈E(G), but this implies {t0 ; vi2 ; z1 ; z1
; vi1 }G  F3 . This 6nal contradiction completes the proof.

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Acknowledgements The authors thank to an anonymous reader for pointing out a small gap in the proof of Theorem 1. References [1] J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, Macmillan, London and Elsevier, New York, 1976. [2] M.R. Garey, D.S. Johnson, Computers and Intractability: a Guide to the Theory of NP-Completeness. Freemann, San Francisco, 1979. [3] N.V.R. Mahadev, B. Reed, A note on vertex orders for stability number, J. Graph Theory 30 (1999) 113–120. [4] O. Murphy, Lower bounds on the stability number of graphs computed in terms of degrees, Discrete Math. 90 (1991) 207–211.